Answer:
3⁻ⁿ⁺²
Step-by-step explanation:
we need to make both bases the same
9 can be written as 3²
[tex] \frac{3 {}^{n} }{3 {}^{2(n - 1)} } [/tex]
multiply 2(n-1) = 2n - 2
[tex] \frac{3 {}^{n} }{3 {}^{2n - 2} } [/tex]
I'll rewrite as 3ⁿ ÷ 3²ⁿ⁻²
subtract the exponents when dividing numbers with the same base
n - (2n-2) or n - 1(2n-2) imagine the 1 is there
remember that the whole of (2n-2) is being multiplied by -1, so expanding
n - 2n + 2 = -n + 2
final answer: 3⁻ⁿ⁺²
